Question

In his doctoral thesis, L. A. Beckel (University of Minnesota, 1982) studied the social behavior of river otters during the mating season. An important role in the bonding process of river otters is very short periods of social grooming. After extensive observations, Dr. Beckel found that one group of river otters under study had a frequency of initiating grooming of approximately 1.7 for each 10 minutes. Suppose that you are observing river otters for 20 minutes. Let r = 0, 1, 2, ... be a random variable that represents the number of times (in a 20-minute interval) one otter initiates social grooming of another.

(a) Explain why the Poisson distribution would be a good choice for the probability distribution of r.

Frequency of initiating social grooming is a common occurrence. It is reasonable to assume the events are dependent.Frequency of initiating social grooming is a common occurrence. It is reasonable to assume the events are independent.     Frequency of initiating social grooming is a rare occurrence. It is reasonable to assume the events are dependent.Frequency of initiating social grooming is a rare occurrence. It is reasonable to assume the events are independent.

What is ??
? =  

Write out the formula for the probability distribution of the random variable r.
P(r) =  

(b) Find the probabilities that in your 20 minutes of observation, one otter will initiate social grooming four times, five times, and six times. (Round your answers to four decimal places.)

P(4) =
P(5) =
P(6) =

(c) Find the probability that one otter will initiate social grooming four or more times during the 20-minute observation period. (Round your answer to four decimal places.)


(d) Find the probability that one otter will initiate social grooming less than four times during the 20-minute observation period. (Round your answer to four decimal places.)

Answer 1

a)Frequency of initiating social grooming is a rare occurrence. It is reasonable to assume the events are independent.

here for 20 m inutes expected number of times =20*1.7/10=3.4 =

P(r)=e^-3.43.4^r/r!

b)

P(4)=0.1858

P(5)=0.1264

P(6)=0.0716

c) probability that one otter will initiate social grooming four or more times during the 20-minute observation period =P(X>=4)=1-P(X<=3)=1-(P(X=0)+P(X=1)+P(X=2)+P(X=3))=1-0.5584=0.4416

d)

probability that one otter will initiate social grooming less than four times during the 20-minute observation period =P(X<4)=1-P(X>=4)=1-0.4416=0.5584